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Two identical, photocathodes receive light of frequencies $f_{1} a n d f_{2}$ . If the velocities of the photoelectrons (of mass m) coming out are respectively $V_{1} a n d V_{2}$ , then

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$V_{1}^{2} - V_{2}^{2} = \frac{2 h}{m} (f_{1} - f_{2})$

$V_{1} + V_{2} = {[\frac{2 h}{m} (f_{1} + f_{2})]}^{1 / 2}$

$V_{1}^{2} + V_{2}^{2} = \frac{2 h}{m} (f_{1} + f_{2})$

$V_{1} - V_{2} = {[\frac{2 h}{m} (f_{1} - f_{2})]}^{1 / 2}$

answer is A.

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Detailed Solution

We know that, $h f = h f_{o} + \frac{1}{2} m v^{2}$

$\Rightarrow \frac{1}{2} m v^{2} = h f - h f_{o}$

$\Rightarrow m v^{2} = 2 h f - 2 h f_{o}$

$\Rightarrow v^{2} = \frac{2 h f}{m} - \frac{2 h f_{o}}{m}$

$∴ {V_{1}}^{2} = \frac{2 h f_{1}}{m} - \frac{2 h f_{o}}{m}$

$∴ {V_{2}}^{2} = \frac{2 h f_{2}}{m} - \frac{2 h f_{o}}{m}$

${\Rightarrow V_{1}}^{2} - {V_{2}}^{2} = \frac{2 h}{m} (f_{1} - f_{2})$

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